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Theorem Let be a smooth projective curve over and an endomorphism of degree . The eigenvalues of on have norm .

Today, we would like to generalize this to varieties of higher dimension. The obvious guess is

Nontheorem Let be a smooth projective variety, over , of dimension . Let be an endomorphism of of degree . The eigenvalues of on have norm .

This is not a theorem! I believe it is Serre who first figured out how to fix and prove this result. That is the topic of today’s post.

Let’s see some counter-examples to the nontheorem. Take . Let and be endomorphisms of , of degrees and . Then has degree . But the action of on is by the matrix ; with eigenvalues and , not the desired .

We can make a similar counter-example with the product of two elliptic curves. In a slight variation, we can work with an abelian surface that has complex multiplication by some rank number ring which has two nonconjugate embeddings into , and choose some who has different norms under the two embeddings.

The fix here is to require to respect the embedding of into projective space. Specifically, Theorem 1 Let be a smooth projective subvariety of . Let be an endomorphism of , of degree , which restricts to an endomorphism of . The eigenvalues of on have norm .

Maps to projective space are closely related to line bundles, and to Kahler classes. Here are the corresponding variants of Theorem 1.

Theorem 1′ Let be a smooth projective variety over ; let be an ample line bundle on ; let be an endomorphism of such that . Then the eigenvalues of on have norm .

Theorem 1” Let be a compact Kahler manifold; let be the Kahler class; let be an endomorphism of such that . Then the eigenvalues of on have norm .

In the previous post, the key trick was to build a positive definite Hermitian form on , for which the action of was unitary. We will want to do the same thing here. Guided by the theorems above, we see that the definition of our Hermitian form should involve the embedding , or equivalently, the Kahler class . For those who don’t know the Kahler story, you can think of as the image in of the hyperplane class from .

We will build our hermitian form in two stages. Fix smooth and projective (or Kahler) of dimension , and a Kahler class with .

The naive bilinear form

For and in , set

.

This is either Hermitian or anti-Hermitian depending on the parity of . And we have .

Unfortunately, this form is not positive definite. Consider . Here is two dimensional. The form does not effect the pairing on middle cohomology (because ) so this is just the Hermitianization of the ordinary pairing on . In particular, it has matrix , with signature .

The Hodge index theorem

Amazingly, there is a classical theorem which exactly addresses this issue. Consider multiplication by as a map from to . We need:

The Hard Lefschetz Theorem: Multiplication by from to is an injection for .

Let’s see how this works for . Let and be the obvious basis for . Let . The hypothesis that is Kahler means that and are positive. (You can think about ample line bundles instead of Kahler classes if you prefer.) So is spanned by . Similarly, is spanned by the kernel of multiplication by , namely . Recalling that is given by , we see that this form is positive definite on and negative definite on as required.

Finishing the proof

Since rescales , it preserves all of the . And they are all orthogonal for . We have already computed that preserves on , and hence on each .

So, on each , the operator preserves a positive definite Hermitian form, and thus its eigenvalues have norm . The eigenvalues of on are the union of the eigenvalues of on each individual .

QED

The Lefschetz operator

We now give a reformulation of the above discussion, which will be very useful in our future presentation.

Let be the operator adjoint to . So, for , we have and, for , we have . In the string , is the operator going the other way.

Note that we can write the projection operator from to in terms of and . Namely,

Define an inner product on by

Hodge index theorem rephrased The restriction of to is times a positive definite Hermitian form.

The exact form of doesn’t matter. The key point is that we can write it as , where is a certain universal noncommutative polynomial in the operators and . When we discuss the characteristic situation, we will take this formulation as our starting point.

Footnote: You might wonder what the standard Hodge index theorem says, and how it is related to its categorified version. We use lower case ‘s for the dimensions of cohomology groups. From , we see that . Also, we have . From these linear equations, we can solve for in terms of , and thus give a formula for the signature of on in terms of the . In particular, when , this relates the signature of the Poincare pairing on to the Hodge numbers of . It is this last relationship which is ordinarily called the Hodge index theorem.

Hi David,
Great post!
Some questions:
1. Is there a similar result for intersection homology of singular projective varieties?
2. Are there some hypotheses we can add, under which we can strengthen this theorem to say that F acts exactly by q^r/2 on H^r?

I was a bit annoyed reading through Kleiman’s treatment in “Dix Exposés…”; I’m rather unfamiliar with the material and he doesn’t give much motivation for the Standard Conjectures. It’s great to see those ideas coming out naturally (here in the characteristic 0 case).

(EDIT: I wrote this answer assuming that F is the Frobenius. See discussion below for other automorphisms.)

Cohomology generated by algebraic cycles is certainly enough. If Y is projective and irreducible of dimension e than F always acts by q^e on H^{2e}(Y). So, if H^{2e}(X) is spanned by classes of algebraic cycles, then functoriality of cohomology for pushforward means that F acts by q^e on H^{2e}(X).

There is a conjectured converse to this. The Tate conjecture has a consequence that, if F acts identically by q^e on H^{2e}(X), then H^{2e}(X) is spanned by algebraic cycles. This is analogous to the characteristic zero situation where the Hodge conjecture implies that, if H^{2e}(X, C) = H^{e,e}(X), then H^{2e}(X,C) is spanned by algebraic cycles. I believe that even these special cases of Tate and Hodge are not known.

There is a detail here which came up in this MO question. I was speaking of the situation where H^{2e} is spanned by algebraic cycles defined over F_q. If the cycles are only defined over a finite extension of F_q, then the action will be by q^e*zeta, for zeta a root of unity.

i don’t have great suggestions for references. Milne’s expository article on Motives, about a year ago, was what inspired me to finally nail this stuff down. I also spent a lot of time reading Kleiman’s article, which you understandably found frustrating, and Grothendieck’s Standard Conjectures article.

I also remember finding a book in the Berkeley library which explained the Kahler analogue very well, back in 2005. Looking through the Berkeley library catalogue to refresh my memory, I think it might have been Zeta functions: An introduction to algebraic geometry, by Thomas? I see that MIT owns this book, so I’ll take a look later today.

Thomas’s book is definitely the book I remembered: pre-LaTeX, has “zeta function” in the title, starts with number fields and moves gently through first function fields and then complex manifolds. But it only proves the case of curves, although it states the higher dimensional result. I thought the book I read had the proof which I wrote up here. I can only conclude that I was reading two books at the same time and they merged together in my memory.

I reshelved the book in the library, so it will still be there for any of our MIT readers who might be looking for it.

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A group blog by 8 recent Berkeley mathematics Ph.D.'s. Commentary on our own research, other mathematics pursuits, and whatever else we feel like writing about on any given day. Sort of like a seminar, but with (even) more rude commentary from the audience.